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olga_2 [115]
3 years ago
15

For hot vacuum filtration, the filter paper should be completely dry when pouring the hot solution into the Buchner funnel to fi

lter.
A) True
B) False
Chemistry
1 answer:
Oliga [24]3 years ago
5 0

Answer:

False.

Explanation:

The given statement is false because for hot vacuum filtration, the filter paper should be wet rather than dry when pouring the hot solution into the Buchner funnel. This is because The possible explanation the filter paper needs to be wetted is not only to allow it to adhere to the funnel, but also to promote the solute to filter down readily across its pores of the paper without wetting it (this is true for organic and aqueous solvents).

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What mass of water is produced by the combustion 1.2 x 10 exp 26 molecules of butane
algol [13]

Answer:

<em>17.94 kg of water</em>

Explanation:

The combustion reaction of balanced butane (C₄H₁₀) is as follows:

2 C₄H₁₀ + 13 O₂ ⇒ 8 CO₂ + 10 H₂O

The amount of butane molecules can be transferred to moles to establish the stoichiometric relationship with water and calculate the moles of water that are formed:

6.02x10²³ molecules _____ 1 mol of butane

1.2x10²⁶ molecules _____ X = 199.34 moles of butane

<em>Calculation</em>: 1.2x10²⁶ molecules x 1 mol / 6.02x10²³ molecules = 199.34 moles of butane

According to the balanced equation:

2 moles butane _____ 10 moles of water

199.34 moles of butane _____ X = 996.68 moles of water

<em>Calculation</em>: 199.34 moles x 10 moles / 2 moles = 996.68 moles of water

And now, you can calculate the mass of the amount of moles obtained from water:

1 mole of water _____ 18 g

996.68 moles of water _____ X = <em>17.94 kg of water </em>

<em>Calculation</em>: 996.68 moles x 18g / 1 mole = 17940.20g ≡ 17.94 kg of water

Therefore, <em>the combustion of 1.2x10²⁶ butane molecules produces 17.94 kg of water</em>.

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2.98 moles of H2 at 35°C and 2.3 atm are in a 32.8 L container. How many moles of H2 are in a 45.3 L container under the same co
zalisa [80]

Answer:

4.12 moles

Explanation:

We can solve this problem with the Ideal Gases Law.

P . V = n . R . T

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P = 2.3 atm

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n = 2.98 moles

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Let's replace data for the second case:

2.3 atm . 45.3L = n . 0.082 . 308K

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